Pyth.計到AE △AHF~△ABE =>計到AH △AJH~△ABE=>計到AJ (AJ/6=AH/AE) 計到JH Let angle AHJ be x angleJAH=90-x angle GHI=180-90-x=90-x △AJH~△HIG HI =6-JH =>計到HI ~△s corr.ratio =>計到GI 有GI同HI ,計到△GIH area IC=JB=6-AJ (計到IC)=>計到HICE area

另外唔識做個時可以自救,又見到幅圖四四方方 introduce coordinate to the geom, let B be (0,0) A(0,6) ,C(6.0) ,F(0,2) 計到AE slope =-6/3=-2 =>FG slope =1/2 寫到equation of AE 同equation of FG 計機solve 相交點H coord. 再put x=6 落equation of FG 可以計到y- coordinate of G ,（即係CG） 有齊H ,G,C,E 四點coordinate ,就計到required area (理論上已經有齊哂題目中所有點嘅coordinate, 乜area 都計到）